Two blocks A and B of equal masses are slidin | vedclass

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Question

(A) The acceleration of a block sliding down an inclined plane is given by $a = g(\sin 	heta - \mu \cos 	heta)$.For block $A$: $a_A = 10(\sin 45^{\c

Vedclass · Accepted Answer

$2 	ext{ s}, 8\sqrt{2} 	ext{ m}$

Vedclass · Answer

(A) The acceleration of a block sliding down an inclined plane is given by $a = g(\sin 	heta - \mu \cos 	heta)$.For block $A$: $a_A = 10(\sin 45^{\circ} - 0.2 \cos 45^{\circ}) = 10 	imes \frac{1}{\sqrt{2}}(1 - 0.2) = \frac{8}{\sqrt{2}} 	ext{ m/s}^2$.For block $B$: $a_B = 10(\sin 45^{\circ} - 0.3 \cos 45^{\circ}) = 10 	imes \frac{1}{\sqrt{2}}(1 - 0.3) = \frac{7}{\sqrt{2}} 	ext{ m/s}^2$.The relative acceleration of $A$ with respect to $B$ is $a_{AB} = a_A - a_B = \frac{8}{\sqrt{2}} - \frac{7}{\sqrt{2}} = \frac{1}{\sqrt{2}} 	ext{ m/s}^2$.Given the initial separation $s_{AB} = \sqrt{2} 	ext{ m}$,we use the equation of motion $s_{AB} = u_{AB}t + \frac{1}{2}a_{AB}t^2$. Since $u_{AB} = 0$,we have $\sqrt{2} = \frac{1}{2} 	imes \frac{1}{\sqrt{2}} 	imes t^2$.Solving for $t$: $t^2 = 4 \Rightarrow t = 2 	ext{ s}$.Now,the distance traveled by block $A$ from its initial position is $s_A = \frac{1}{2}a_A t^2 = \frac{1}{2} 	imes \frac{8}{\sqrt{2}} 	imes (2)^2 = \frac{1}{2} 	imes \frac{8}{\sqrt{2}} 	imes 4 = \frac{16}{\sqrt{2}} = 8\sqrt{2} 	ext{ m}$.

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